What is the domain of the function \(f(x)=\frac{\sqrt{25-x^2}}{x-3}\)?

Correct answer: \([-5,5]\setminus\{3\}\)

The expression under the square root must be non-negative, so \(25-x^2\ge 0\), which gives \(-5\le x\le 5\). Thus the initial domain is \([-5,5]\). Also, the denominator \(x-3\) cannot be zero, so \(x=3\) must be excluded. Therefore, the domain is \([-5,5]\setminus\{3\}\). Option B is incorrect because the endpoints \(-5\) and \(5\) are allowed. Exam tip: impose \(\text{radicand}\ge 0\) and denominator \(\ne 0\).